Switching mode power supplies are widely used in various electronic devices for providing stable voltages; however, non-ideal effects resulted from the switches, inductor and output capacitor in a switching mode power supply may cause the output voltage of the switching mode power supply unstable, and thus a compensator is needed to compensate these non-ideal effects. FIG. 1 shows a conventional switching mode power supply 100, which includes an output stage 102 to convert an input voltage Vi to an output voltage Vo for a load 104, a pre-filter 105 to filter the output voltage Vo to produce a voltage Vo′, an error amplifier 106 to compare the voltage Vo′ with a reference voltage Vref to generate an error signal Ve, a compensator 108 to compensate the error signal Ve to generate a signal Vd, and a pulse width modulator to generate a pulse width modulation (PWM) signal pwm_o according to the signal Vd to drive the output stage 102. In the output stage 102, a driver 112 is used to switch a pair of switches 114 and 116 connected in series between the power input Vi and ground GND according to the PWM signal pwm_o, in order to convert the input voltage Vi to the output voltage Vo.
Generally speaking, the compensator 108 for the switching mode power supply 100 is designed based on the small signal modeling of the switching mode power supply 100. The continuous-time transfer function of each circuit can be derived from the small signal analysis and Laplace transformation as shown in FIG. 2, in which the pre-filter 105 has the transfer function H(s) to transform the output voltage {circumflex over (V)}o(s) into the voltage {circumflex over (V)}o′(s)=H(s)×{circumflex over (V)}o(s), the error amplifier 106 compares the voltage {circumflex over (V)}o′(s) with a reference voltage {circumflex over (V)}ref(s) to generate the error signal {circumflex over (V)}e(s), the compensator 108 has the transfer function Gc(s) to compensate the error signal {circumflex over (V)}e(s) to generate the signal {circumflex over (V)}d(s), the pulse width modulator 110 has the transfer function 1/VM to transform the signal {circumflex over (V)}d(s) into the PWM signal pwm_o(s), and the output stage 102 has the transfer function Gvd(s) of the so-called control-to-output voltage to transform the PWM signal pwm_o(s) into the output voltage {circumflex over (V)}o(s). From the system shown in FIG. 2, the switching mode power supply 100 obviously has the open-loop gainT(s)=Gvd(s)×Gc(s)×H(s)/VM.  [Eq-1]Further, the continuous-time transfer function of the output stage 102 is
                                                                        Gvd                ⁡                                  (                  s                  )                                            =                                                V                  ^                                ⁢                                                      o                    ⁡                                          (                      s                      )                                                        /                  pwm_o                                ⁢                                  (                  s                  )                                                                                                        =                                                                    (                                          RoRcCs                      +                      Ro                                        )                                    ⁢                  Vi                                                                                                                                                                          LC                            ⁡                                                          (                                                              Rc                                +                                Ro                                                            )                                                                                ⁢                                                      s                            2                                                                          +                                                                                                                                                                                                      [                                                          L                              +                                                                                                (                                                                      RoRc                                    +                                                                          RcR                                      L                                                                        +                                                                          RoR                                      L                                                                                                        )                                                                ⁢                                C                                                                                      ]                                                    ⁢                          s                                                +                                                  (                                                                                    R                              L                                                        +                            Ro                                                    )                                                                                                                                                                                            =                                                (                                                                                    b                        1                                            ⁢                      s                                        +                                          b                      0                                                        )                                /                                  (                                                                                    a                        2                                            ⁢                                              s                        2                                                              +                                                                  a                        1                                            ⁢                      s                                        +                                          a                      0                                                        )                                                                                                                        =                                                      A                    ⁡                                          [                                                                        (                                                      s                            /                                                          ω                                                              z                                ⁢                                                                                                                                  ⁢                                1                                                                                                              )                                                +                        1                                            ]                                                        /                                      {                                                                  [                                                                              (                                                          s                              /                                                              ω                                                                  p                                  ⁢                                                                                                                                          ⁢                                  1                                                                                                                      )                                                    +                          1                                                ]                                            ⁡                                              [                                                                              (                                                          s                              /                                                              ω                                                                  p                                  ⁢                                                                                                                                          ⁢                                  2                                                                                                                      )                                                    +                          1                                                ]                                                              }                                                              ,                                                          [                  Eq          ⁢                      -                    ⁢          2                ]            where a0, a1, a2, b0, b1 and A are constants, ωp1 and ωp2 are poles, ωz1 is a zero, RL is the equivalent DC resistance of the inductor L, Rc is the equivalent serial resistance of the capacitor C, and Ro is the resistance of the load 104. From the equation Eq-2, it can be obtained the poles
                                          ω                          p              ⁢                                                          ⁢              1                                =                                                                      a                  1                                                  a                  2                                            +                                                                                          (                                                                        a                          1                                                                          a                          2                                                                    )                                        2                                    -                                      4                    ⁢                                                                  a                        0                                                                    a                        2                                                                                                                  2                          ⁢                                  ⁢        and                            [                  Eq          ⁢                      -                    ⁢                                          ⁢          3                ]                                                      ω                          p              ⁢                                                          ⁢              2                                =                                                                      a                  1                                                  a                  2                                            -                                                                                          (                                                                        a                          1                                                                          a                          2                                                                    )                                        2                                    -                                      4                    ⁢                                                                  a                        0                                                                    a                        2                                                                                                                  2                          ,                            [                  Eq          ⁢                      -                    ⁢          4                ]            the zeroωz1=b0/b1,  [Eq-5]and the constantA=(b0a2)/(a1a0).  [Eq-6]
The compensator 108 is so designed according to the transfer function Gvd(s) of the output stage 102 for the converter 100 to provide a stable output voltage and a fast load transient response. Currently, popular analog compensators include type-I, type-II and type-III, and FIG. 3 shows a typical type-III analog compensator 108 which includes resistors R1, R2, R3, and capacitors C1, C2, C3. From the circuit of FIG. 3, the analog compensator 108 has the continuous-time transfer function
                                                                        Gc                ⁡                                  (                  s                  )                                            =                                                                    (                                          1                      +                                              sR                        ⁢                                                                                                  ⁢                        1                        ⁢                                                                                                  ⁢                        C                        ⁢                                                                                                  ⁢                        2                                                              )                                    ⁡                                      [                                          1                      +                                                                        s                          ⁡                                                      (                                                                                          R                                ⁢                                                                                                                                  ⁢                                3                                                            +                                                              R                                ⁢                                                                                                                                  ⁢                                1                                                                                      )                                                                          ⁢                        C                        ⁢                                                                                                  ⁢                        3                                                              ]                                                                    sR                  ⁢                                                                          ⁢                  1                  ⁢                                      (                                          1                      +                                              sC                        ⁢                                                                                                  ⁢                        3                        ⁢                                                                                                  ⁢                        R                        ⁢                                                                                                  ⁢                        3                                                              )                                    ⁢                                      (                                                                  C                        ⁢                                                                                                  ⁢                        1                                            +                                              C                        ⁢                                                                                                  ⁢                        2                                            +                                              sR                        ⁢                                                                                                  ⁢                        2                        ⁢                        C                        ⁢                                                                                                  ⁢                        1                        ⁢                        C                        ⁢                                                                                                  ⁢                        2                                                              )                                                                                                                                          =                                  Ac                  ⁢                                                                                    (                                                  1                          +                                                      s                            /                                                          ω                                                              cz                                ⁢                                                                                                                                  ⁢                                1                                                                                                                                    )                                            ⁢                                              (                                                  1                          +                                                      s                            /                                                          ω                                                              cz                                ⁢                                                                                                                                  ⁢                                2                                                                                                                                    )                                                                                                            s                        ⁡                                                  (                                                      1                            +                                                          s                              /                                                              ω                                                                  cp                                  ⁢                                                                                                                                          ⁢                                  1                                                                                                                                              )                                                                    ⁢                                              (                                                  1                          +                                                      s                            /                                                          ω                                                              cp                                ⁢                                                                                                                                  ⁢                                2                                                                                                                                    )                                                                                                        ,                                                          [                  Eq          ⁢                      -                    ⁢          7                ]            where Ac is a gain constant, ωcz1 and ωcz2 are zeros, and ωcp1 and ωcp2 are poles. In the equation Eq-7, the poles ωcp1 and ωcp2 are real poles for suppressing high-frequency noises. A good compensator 108 must eliminate the poles in the denominator of the transfer function Gvd(s) of the output stage 102. However, due to the inductor L and the output capacitor C, the transfer function Gvd(s) usually includes complex conjugated pole pair or a right hand zero (RHZ). For example, when
            (                        a          1                          a          2                    )        2    <      4    ⁢                  a        0                    a        2            in the equations Eq-3 and Eq-4, the poles ωp1 and ωp2 become complex conjugated pole pair. As shown in the equation Eq-7, the poles ωcp1 and ωcp2 of the analog compensator 108 are
            1              C        ⁢                                  ⁢        3        ⁢        R        ⁢                                  ⁢        3              ⁢                  ⁢    and    ⁢                  ⁢          1              R        ⁢                                  ⁢        2        ⁢                                  ⁢        C        ⁢                                  ⁢        1        ⁢        C        ⁢                                  ⁢        2              ,respectively. Since R2, R3, C1, C2 and C3 are all real, the poles ωp1 and ωp2 are real and thus cannot produce a complex number. As a result, the complex conjugated pole pair in the transfer function Gvd(s) of the output stage 102 will not be eliminated. In other words, the traditional compensator 108 can only eliminate the magnitude term of the complex conjugated pole pair so as to compensate the gain, but cannot compensate the phase angle correctly, which will reduce the bandwidth and the phase margin of the switching mode power supply. In addition, current digital compensators only compensate the magnitude term of the complex conjugated pole pair.
For more information of analog compensators, readers are referred to R. W. Erickon and D. Maksimoric, “Fundamentals of Power Electronics”, Kluwer Academic Publisher, 2001. For more information of digital compensators, readers are referred to D. Maksimovic, R. Zane and R. Erickson, “Impact of Digital Control in Power Electronics”, Proceeding of 2004 International Symposium on Power Semiconductor Devices & Ics, pp. 13-22, and Shamim Choudhury, “Designing a TMS320F280x Based Digitally Controlled DC-DC Switching Power Supply”, Texas Instruments Application Report, SPRAAB3, July, 2005.
For further details, a unit gain single pole system is taken as an example which has the transfer functionI(s)=1/[1+(s/ωp)].  [Eq-8]The equation Eq-8 means that the system has a pole at the angular frequency ωp, and the poleωp=ωp0ejθ=ωp0(cos θ+j sin θ),  [Eq-9]where ωp0 is the magnitude of the pole ωp, and θ is the phase angle of the pole ωp. From the equation Eq-9, when sin θ=0, for example θ=N×180, where N is an integer, ωp is a real pole. When sin θ≠0, ωp is a complex pole. Obviously, even having a same magnitude ωp0, the complex pole ωp will have different frequency response with different phase angles θ. FIGS. 4 and 5 show the frequency response of the unit gain single pole system at different phase angles. In FIG. 4, X-axis represents the frequency, Y-axis represents the gain, curve 200 represents the gain when the phase angle θ of the system is zero, curve 202 represents the gain when the phase angle θ of the system is 45 degrees, curve 204 represents the gain when the phase angle θ of the system is 90 degrees, and curve 206 represent the gain when the phase angle θ of the system is 135 degrees. In FIG. 5, X-axis represents the frequency, Y-axis represents the phase, curve 208 represents the phase when the phase angle θ of the system is zero, curve 210 represents the phase when the phase angle θ of the system is 45 degrees, curve 212 represents the phase when the phase angle θ of the system is 90 degrees, and curve 214 represents the phase when the phase angle θ of the system is 135 degrees. Let ωp0=2π×1000 and θ=0, the unit gain single pole system has the transfer functionI(s)=1/[1+(s/(2π×1000))].  [Eq-10]This means that when the pole is at the frequency 1,000 Hz, as shown by the curve 200 in FIG. 4, the unit gain single pole system is a real pole system. When the frequency is higher than 10 KHz, the phase of the system is close to −90 degrees, as shown by the curve 208 in FIG. 5. However, if the pole ωp has a non-zero phase angle, for example 45 degrees, the transfer function of the system will become
                                                                        I                ⁡                                  (                  s                  )                                            =                              1                /                                  [                                      1                    +                                          (                                              s                        /                                                  (                                                      2                            ⁢                            π                            ×                            1000                            ⁢                                                                                                                  ⁢                                                          ⅇ                                                              in                                /                                4                                                                                                              )                                                                    )                                                        ]                                                                                                        =                              1                /                                                      [                                          1                      +                                              (                                                  s                          /                                                      (                                                          2                              ⁢                              π                              ×                              1000                              ⁢                                                              (                                                                  0.7071                                  +                                                                      0.7071                                    ⁢                                                                                                                                                  ⁢                                    i                                                                                                  )                                                                                      )                                                                          )                                                              ]                                    .                                                                                        [                  Eq          ⁢                      -                    ⁢          11                ]            From the equation Eq-11, as the system has an imaginary number 2π×1000×0.7071i, the phase of the system is close to −45 degrees when the frequency is over 10 KHz. As shown by the curve 212 in FIG. 5, it is obvious that the phase of the system is close to −45 degrees at high frequencies, but not close to 90 degrees as the situation when the system is at a single real pole. FIGS. 6 and 7 show the difference between a perfect and an imperfect compensated system. In FIG. 6, X-axis represents the frequency, Y-axis represents the gain, curves 200 and 204 are the gain when the phase angle θ is zero and 45 degrees, respectively, curve 220 represents the gain after the pole of the system is perfect compensated, and curve 222 represents the gain after the complex pole is compensated by a real zero. In FIG. 7, X-axis represents the frequency, Y-axis represents the phase, curves 208 and 212 are the phase when the phase angle θ is zero and 45 degrees, respectively, curve 224 represents the phase after the pole of the system is perfect compensated, and curve 226 represents the phase after the complex pole is compensated by a real zero. When the system has a single pole, if there is a compensator C(s) which can provide only one real zero, for example C(s)=(1+s/(2π×1000)), this compensator will compensate a real pole system to have the ideal characteristic of the gain equal to 1 and the phase equal to 0 as indicated by the equation Eq-10, as shown by the curves 220 and 224 in FIGS. 6 and 7. However, when a system having complex conjugated pole pair as the equation Eq-11 indicates is compensated by the compensator C(s), it is obtained
                                          I            ⁡                          (              s              )                                ·                      C            ⁡                          (              s              )                                      =                                            1              +                              s                                  2                  ⁢                  π                  ×                  1000                                                                    1              +                              s                                  2                  ⁢                  π                  ×                  1000                  ⁢                                                                          ⁢                                      ⅇ                                          ⅈπ                      4                                                                                                    .                                    [                  Eq          ⁢                      -                    ⁢          12                ]            The frequency response is shown by the curves 222 and 226 in FIGS. 6 and 7, and the system has the non-ideal characteristic of the gain not equal to 1 and the phase not equal to 0.
Therefore, there is a need of a compensator for eliminating the complex conjugated pole pair of a switching mode power supply.